# Prove that $H$ is the orthocenter of $\Delta ABC$.

In acute $\Delta ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively such that $B, C, D, E$ are concyclic, $BE$ and $CD$ intersect at $H$, and $H$ is on the altitude of $\Delta ABC$ passing through $A$. Prove that $H$ is the orthocenter of $\Delta ABC$.

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Is this true? For $H$ to be the orthocenter, then line $CH$ must be perpendicular to $AB$; but line $CH$ is line $CD$, so that $D$ must be the foot of the perpendicular from $C$: that is, it cannot be chosen arbitrarily. On the other hand, there are plenty of choices of concyclic points $B$, $C$, $D$, $E$. –  Blue Feb 3 '13 at 3:45
Or consider this: By taking $D$ and $E$ sufficiently close to vertex $A$, the point $H$ gets arbitrarily close to $A$, which is (usually) not the orthocenter. But just the fact that you can move $H$ around with choices of $D$ and $E$ shows that it can't be the fixed orthocenter. –  Blue Feb 3 '13 at 3:49
@Blue I think you should provide an answer to this problem. It's quite clear that this is not true. –  EuYu Feb 3 '13 at 3:51
@jasoncube That doesn't change anything. The result doesn't hold for acute triangles either. –  EuYu Feb 3 '13 at 3:56
Sorry, I've just missed an important information. –  ᴊ ᴀ s ᴏ ɴ Feb 3 '13 at 4:00

Theorem: Let $O$ be a circle with chord $AB$. Let $\widehat{AB}$ denote an arc of $O$ ending on $A$ and $B$. Choose $C$ and $D$ on $\widehat{AB}$ such that $C$ is adjacent to $B$ (and $D$ is adjacent to $A$) and let $AD$ intersect $BC$ at $K$. Let ${AC}$ intersect $BD$ at $H$. Then $H$ lies on the perpendicular of $AB$ through $K$ if and only if $AB$ is a diameter of $O$ or $AK = BK$.

Proof: Suppose the perpendicular of $AB$ through $K$ intersects $AB$ at $G$. Then $KG$, $AC$ and $BD$ are cevians of the triangle. By the trigonometric form of Ceva's theorem, the three are concurrent if and only if $$\sin(\angle KBD)\sin(\angle BAC)\sin(\angle AKG) = \sin(\angle ABD)\sin(\angle KAC)\sin(\angle BKG)$$ Let us now compile some angular equivalencies present in $\triangle ABK$. First, note that $$\angle KAC = \angle KBD$$ since they both subtend $\widehat{CD}$. By a bit of angle chasing, we also have $$\angle ABD = (90 - \angle ADB) + \angle AKG$$ $$\angle BAC = (90 - \angle ACB) + \angle BKG$$ Again, notice that $\angle ADB = \angle ACB$ since they both subtend $\widehat{AB}$. Let $x$ denote $90 - \angle ACB$. From this, Ceva's theorem simplifies as $$\sin(x + \angle BKG)\sin(\angle AKG) = \sin(x + \angle AKG)\sin(\angle BKG)$$ expanding the sines using angle addition identities, the equation reduces to $$\sin(x)\cos(\angle BKG)\sin(\angle AKG) = \sin(x)\cos(\angle AKG)\sin(\angle BKG)$$ or equivalently $$\sin(x)\tan(\angle AKG) = \sin(x)\tan(\angle BKG)$$ clearly we have equality if and only if $\sin(x) = 0$ which implies $$\angle ACB = \angle ADB = 90$$ This of course implies that $AB$ is a diameter of $O$. Or we have $$\tan(\angle AKG) = \tan(\angle BKG) \implies \angle AKG = \angle BKG$$ in which case $\triangle AKB$ is isosceles with $AK = BK$. $\square$

Now, suppose that the hypothesis of your proposition is satisfied. Let $O$ denote the circumcircle of $BCDE$ of which $BC$ is a chord. Since $H$ lies on the altitude through $A$, by the above theorem, either triangle $\triangle ABC$ is isosceles or $BC$ is a diameter of $O$. If the latter, then $CD \perp AB$ and $BE \perp AC$ so that $H$ is indeed the orthocenter of the triangle. Note that the condition that $\triangle ABC$ is acute is not needed.

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The (final) statement is not true.

Consider an acute isosceles triangle with vertex angle $A$. Because the figure is symmetric about the altitude through $A$, any circle through $B$ and $C$ that meets $AB$ and $AC$ will meet these edges at points $D$ and $E$ such that $BD$ and $CE$ have intersection $H$ somewhere on that altitude.

On one extreme, the smallest circle under consideration is the one with diameter $BC$, which in fact leads to orthocenter $H$ (since $\angle BDC$ and $\angle BEC$ subtend a semicircle); on the other, the largest circle is the circumcircle of $\triangle ABC$, for which we have $H=A$. Other circles lead to points in between the orthocenter and $A$, so the conditions do not "usually" cause $H$ to be the orthocenter.

A note about the acuteness condition: were $\triangle ABC$ a right isosceles triangle, the two "extreme" circles mentioned above would coincide, and we would indeed have $H=A$ lie at the orthocenter. Were the triangle obtuse, then it would lie entirely within the smallest circle (with diameter $BC$), allowing no intersection points $D$ and $E$ with the sides of the triangle. (Extended sides are another matter.)

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